Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

by
Knight, Randall D.

Answer

(a) $t = 0.064~s$
(b) The bullet's speed as it left the barrel was 780 m/s

Work Step by Step

(a) We can find the flight time of the bullet:
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(0.020~m)}{9.80~m/s^2}}$
$t = 0.064~s$
(b) We can find the bullet's speed as it left the barrel;
$v_x = \frac{x}{t} = \frac{50~m}{0.064~s}$
$v_x = 780~m/s$
The bullet's speed as it left the barrel was 780 m/s.